Use calculus to approximate e^0.1

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Күн бұрын

Use calculus, NOT calculators! We will use a tangent line approximation and differentials to approximate e^0.1. This is called local linear approximation which is an application of derivatives in calculus 1. Subscribe for more calculus tutorials 👉 bit.ly/just_calc
0:00 use a local linear approximation for e^0.1
3:46 use differentials to approximate e^0.1
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Пікірлер: 56

@abdulllllahhh 2 жыл бұрын

Just aced my calc 2 test at school thanks to you. Love you BPRP

@Purplesjh 2 жыл бұрын

Great job! May I ask which kind of problems were there on the test?

@bprpcalculusbasics 2 жыл бұрын

Glad to hear! Well done!

@captainhd9741 2 жыл бұрын

@@bprpcalculusbasics make videos about P.D.E.s

@Gauravsingh-zg1lc 2 жыл бұрын

Also can use maclaurin series of e^x to approximate it.

@gurkiratsingh7tha993 2 жыл бұрын

I agree, this is a good Approximation

@vasnormandy2 2 жыл бұрын

And getting the third term nets you 1.105 which is even closer ^^

@volodymyrgandzhuk361 2 жыл бұрын

But it takes longer (unless you have practiced it enough)

@aldobernaltvbernal8745 2 жыл бұрын

@@volodymyrgandzhuk361 not really the input is 0.1 so you have 1 + 0.1 + 0.01/2 + 0.001/6 + ... and just taking the first 3 terms you can do 1.1 + 0.01/2 and since you have a 0.01 on the numerator that's kinda like shifting the decimal point in 0.5 so your answer is 1.105 and i'd argue you can do that much quicker than with the method done in the video, and also get closer.

@volodymyrgandzhuk361 2 жыл бұрын

@@aldobernaltvbernal8745 I didn't mean for this particular case

@ZipplyZane 2 жыл бұрын

Aren't both approaches doing the same thing "under the hood"? I like version 1 to see why it works as an approximation. But I like version 2 because we then see how it gives value to dy and dx.

@AntonFediukov 2 жыл бұрын

Hey BPRP, you should make a video on the taylor series of e^(x+1) centered at 0, you can use exponent rules to show that its just e times the maclaurin series of e^x or you can use a double summation with the binomial expansion to get the same result with some pretty funky indexing. Love your videos!

@josevidal354 2 жыл бұрын

Some time ago I Made little formula to aproach a^x and log_a(x). They are these: a^x ≈ a^floor(x)*[1+(a-1)(x-floor(x))] log_a(x) ≈ floor(log_a(x))+[x*a^-floor(log_a(x))-1]/(a-1) They may seem complicated but when they are graphed they are just a bunch of lines that connect every unit and are touching the function

@bondarakhanganeh 2 жыл бұрын

exp(0.1)~exp(0)=1, so exp(0.1)~1

@TheBodyOnPC 2 жыл бұрын

e^r~1+r for small r

@factorsistemas5396 2 жыл бұрын

You Sir are amazing , thank you so much from Brazil

@yograjbir9993 2 жыл бұрын

In india , we prefer second method because it is just cool

@silentintegrals9104 2 жыл бұрын

Nice video!!

@nisa.hlfmntn 2 жыл бұрын

thank you sooooo much

@gurkiratsingh7tha993 2 жыл бұрын

Next time, write all the values of the zeta function in terms of pi

@pardeepgarg2640 2 жыл бұрын

Did u done with that integral Of writing Zeta(a+bi) in terms of integral U posts that cmnt months ago?

@gurkiratsingh7tha993 2 жыл бұрын

@@pardeepgarg2640 yes, it was not easy but i did it

@ahmedabuarkoub648 2 жыл бұрын

Kabby lame be like e^0=1 e^0.1 1+0.1 so it 1.1

@bprpcalculusbasics 2 жыл бұрын

Can't agree more! lol

@theimmux3034 2 жыл бұрын

You can get a better approximation by setting the secant approximation of a slope at a point equal to the slope at that point. Say you want to find the slope at x = 0,1/2. Then: e^(0,1/2) (approx.)= (e^(0,1) - e^0)/(0,1-0) This leads to a quadratic in terms of e^(0,1/2) e^(0,1/2) = (1 + sqrt(401))/20 Square both sides and you get: e^0,1 = (201 + sqrt(401))/200 = 1,105124921... The error is about 4,16·10^(-5)

@reeeeeplease1178 2 жыл бұрын

Problem being that you still need a calculator, tho its nice

@reeeeeplease1178 2 жыл бұрын

And where does the sqrt come from

@theimmux3034 2 жыл бұрын

@@reeeeeplease1178 You solve a quadratic. And you don't need a calculator.

@kinshuksinghania4289 2 жыл бұрын

1.1 using linear approximation from point (0,1) and f'(0) = 1

@martinepstein9826 2 жыл бұрын

Man, that all looks so complicated to me. rise = slope * run. We start at y = e^0 = 1. The slope is the derivative of e^x at 0 which is 1. The run is 0.1. e^0.1 ~= 1 + 1*0.1 = 1.1

@martinepstein9826 2 жыл бұрын

f(x+h) = f(x) + f'(x)h + o(h) e^(0+0.1) = e^0 + 1*0.1 + o(0.1) ~= 1.1

@zoltankurti 2 жыл бұрын

@@martinepstein9826 +o(h^2)...

@martinepstein9826 2 жыл бұрын

@@zoltankurti Nope, o(h). "Small relative to h". You might be thinking of O(h^2), "comparable to h^2 (or smaller)".

@captainhd9741 2 жыл бұрын

Do partial differential equations

@silentintegrals9104 2 жыл бұрын

totally agree

@numberandfacts6174 2 жыл бұрын

What is answer for s if Γ(s) = ∞

@materiasacra 2 жыл бұрын

Glance at the graph of the Gamma function (en.wikipedia.org/wiki/File:Gamma_plot.svg) and you get a pretty good idea of the answer. Maybe that is enough. If not, you can consider the integral representation \Gamma(z) = \int _{0}^{\infty } x^{z-1} e^{-x} dx , which is usually employed to define the gamma function for non-natural z. This converges at x=0 for all Re(z)>0. It clearly diverges for z=0 (like integral 1/x at x=0). You can use integration by parts to establish the property \Gamma(z+1) = z \Gamma(z) for Re(z)>0, which is very useful for sideways stepping in the complex z-plane. This property also holds for the analytic continuation into the area where Re(z)

@xinpingdonohoe3978 2 жыл бұрын

s=∞ Or rather, s=lim(t→∞)(t) Assuming you mean gamma as in factorial.